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Need help on Electrical Field problem

  1. Nov 19, 2004 #1
    Four charges are at the corners of a square of side a as show in the figure. a) Determine the magnitude and direction of the electric field at the location of charge q. b) What is the resultant force on q?
    Code (Text):
    http://east.ilrn.com/graphing/bca/user/appletImage?dbid=1131113015
    I made E1 equal the electric field of 2q and q. I made E2 equal the electric field of 3q and q. I made E3 equal the electric field of 4q and q. So, when I wrote out the vector notations of each electric field, I got:
    [tex] E_{1} = k_{e} \frac {2q} {a^2} \hat {i} [/tex]
    [tex] E_{3} = k_{e} \frac {4q} {a^2} \hat {j} [/tex]
    [tex] E_{2} = k_{e} \frac {3q} {a^2} [/tex]
    How would u break down E2 into x and y components?

    Also, after getting E1, E2, and E3, E = E1 + E2 + E3. So, you add all the vectors up and a x and y component. Next, magnitude of E = square root of x component squared plus y component squared. The angle is tan of y component over x component.

    Did I do it right for part A? Any help would be great! thx !!
     
  2. jcsd
  3. Nov 19, 2004 #2

    HallsofIvy

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    I'm not sure what you mean by "electric field of 2q and q" since you don't give any electric field, just the force between 2q and q. Even for those, I don't see how you got the "2q", "4q", and "3q" in the numerators. The force between two charges depends upon the product of their charges. Assuming that "2q" means the charge is twice that of q, etc. the forces are
    [tex] F_{1} = k_{e} \frac {2q^2} {a^2} \hat {i} [/tex]
    [tex] F_{2} = k_{e} \frac {3q} {a^2} [/tex]

    You also seem missing the fact that the distance between "q" and "3q" is √(2)a, not a.
    [tex] F_{3} = k_{e} \frac {2q^2} {2a^2} \hat {j} [/tex]

    Those are the magnitudes. Of course F1 has positive x component and 0 y component while F3 has positive y component and 0 x component.

    F2 has equal x and y components, equal to the magnitude of F2 above multiplied by 1/√(2).
     
  4. Nov 19, 2004 #3

    Andrew Mason

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    Your expression for E2 is incorrect. The separation is [itex]\sqrt{2a^2}[/itex]

    Resolve E2 into its [itex]\hat{i}[/itex] and [itex]\hat{j}[/itex] components and add to E1 and E3 respectively.That gives you the orthogonal components of the resulting vector. Just work out the direction and magnitude from those components to get the resulting field. Then use [itex]\vec{F} = q\vec{E}[/itex]

    AM
     
  5. Nov 19, 2004 #4
    thx a lot!
     
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